A stone of mass $1 k g$ is tied to the end of a string of $1 m$ long. It is whirled in a vertical circle. If the velocity of the stone at the top be $4 m / s$ . What is the tension in the string?

6 N

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16 N

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5 N

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10 N

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Solution

The correct option is A $6 N$
Given : $v = 4$ $m / s$ $m = 1 k g$ $r = 1$ m
Using circular motion equation : $\frac{m v^{2}}{r} = T + m g$
$⟹$ $T = \frac{m v^{2}}{r} - m g$

$∴$ $T = \frac{1 \times 4^{2}}{1} - 1 \times 10$ $⟹ T = 6$ $N$

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A stone of mass 1 kg is tied with a string and it is whirled in a vertical circle of radius 1m. If tension at the highest point is 14N, then velocity at lowest point will be:

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A stone of mass 1kg is tied to the end of a string of 1m long. It is whirled in a vertical circle. If the velocity of the stone at the top be 4m/s. What is the tension in the string?

The correct option is A 6NGiven : v=4 m/s m=1kg r=1 mUsing circular motion equation : mv2r=T+mg⟹ T=mv2r−mg∴ T=1×421−1×10 ⟹T=6 N

A stone of mass $1 k g$ is tied to the end of a string of $1 m$ long. It is whirled in a vertical circle. If the velocity of the stone at the top be $4 m / s$ . What is the tension in the string?

The correct option is A $6 N$
Given : $v = 4$ $m / s$ $m = 1 k g$ $r = 1$ m
Using circular motion equation : $\frac{m v^{2}}{r} = T + m g$
$⟹$ $T = \frac{m v^{2}}{r} - m g$

$∴$ $T = \frac{1 \times 4^{2}}{1} - 1 \times 10$ $⟹ T = 6$ $N$